Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
 A: If $ab+c=bc+a$ then $ab-bc+c-a=0$ that is $(a-c)(b-1)=0$.
Thus $a=c$ or $b=1$. Similarly $a=b$ or $c=1$. Also $b=c$ or $a=1$.
Combining these we get various cases to consider. For instance,
if $a=c$ and $a=b$ then $4=a^2+a$ etc. Or if $a=c$ and $c=1$ then
$4=ac+b=1+b$ so $b=3$ etc.
A: If $a b+c=4$ then $a=\frac{4-c}{b}$ assuming that $b\not=0$ and then $b c+a =4$ which is $b c+\frac{4-c}{b}=4$ so $c(b-\frac{1}{b})+\frac{4}{b} = 4$ so $c= \frac{4-\frac{4}{b}} { b-\frac{1}{b}} = \frac{4}{b+1}$ assuming that $b\not -1,1$ and to the final equation we get that $c a+b=4$ which is $c(\frac{4-c}{b})+b=4$ which is $\frac{4}{b+1}(\frac{4-\frac{4}{b+1}}{b})+b=4$ and solving for $b$ we get that $b=3,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so $c=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ and $a=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so the only solutions are :
$(1,3,1)$ and all its permutations and $(\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17})),(\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}))$
A: $ab + c = bc + a = ca + b=4$ 
By symmetry any solution for any any one of $a,b,c$ can be a solution for any other.
$ab +c = bc +a$ means $ab - bc = b(a-c) = a-c$  So either $b=1$ or $a=c$.
So by symmetry:
if $a\ne 1$ then $b=c$ and if $b\ne 1$ then $a = c = b$.  So it is not possible that only one of $a,b,c$ is equal to $1$.
So either:
1) All $a,b,c$ equal $1$
2) None of $a,b,c$ equal $1$ and $a=b=c$.
3) Two of $a,b,c$ equal $1$ and the third does not.
....
1) $1*1 + 1 = 4$ is not possible.
2)we get $a^2 + a = 4$ and $a = \frac {-1 \pm \sqrt{17}}{2}$.
3) wolog $a= b = 1$ so $1 + c = c + 1 = c + 1 = 4$.  $c = 3$.
So there are 5 solutions:
And $(a,b,c) = (1,1,3)$, $(1,3,1)$ , $(3,1,1), ( \frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2)$ or $ ( \frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2)$
